The cross-section of a small river is sub-divided into seven segments of width $1.5 \mathrm{~m}$ each. The average depth, and velocity at different depths were measured during a field campaign at the middle of each segment width. The discharge computed by the velocity area method for the given data is ____ $\mathrm{m}^{3} / \mathrm{s}$ (round off to one decimal place).

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📖 Explanation

Average width for 1 st land last sections,

\begin{aligned} \bar{\omega}_{1}=\bar{\omega}_{\mathrm{n}} & =\frac{\left(\omega_{1}+\frac{\omega_{2}}{2}\right)^{2}}{2 \omega_{1}} \\ & =\frac{\left(1.5+\frac{1.5}{2}\right)^{2}}{2 \times 1.5} \\ & =1.6875 \mathrm{~m} \end{aligned}

\begin{array}{|c|c|c|c|} \hline \begin{array}{l}\text { Avg. width } \\ \text { (m) } \\ \text { (1) }\end{array} & \begin{array}{l}\text { Avg. Depth } \\ \text { (m) } \\ \text { (2) }\end{array} & \begin{array}{l}\text { Avg. Velocity } \\ \text { (m/s) } \\ \text { (3) }\end{array} & \begin{array}{c}\text { Segmental } \\ \text { discharge } \\ \left(\mathrm{m}^{3} / \mathrm{s}\right)=(1) \times (2) \times (3)\end{array} \\ \hline 1.6875 & 0.40 & 0.40 & 0.270 \\ \hline 1.5 & 0.70 & \frac{0.76+0.70}{2}=0.73 & 0.767 \\ \hline 1.5 & 1.20 & \frac{1.19+1.13}{2}=1.16 & 2.088 \\ \hline 1.5 & 1.40 & \frac{1.25+1.10}{2}=1.175 & 2.467 \\ \hline 1.5 & 1.10 & \frac{1.13+1.09}{2}=1.11 & 1.831 \\ \hline 1.5 & 0.80 & \frac{0.69+0.65}{2}=0.67 & 0.804 \\ \hline 1.6875 & 0.45 & 0.42 & 0.319 \\ \hline & & & \Sigma=8.546 \\ \hline \end{array}

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